Problem: Let $f$ be a transformation from $\mathbb{R}^2$ to $\mathbb{R}^2$. Its Jacobian matrix is given below. $J(f) = \begin{bmatrix} -\sin(x) & -\cos(y) \\ \\ \cos(x) & \sin(y) \end{bmatrix}$ Find the Jacobian determinant of $f$. $|J(f)| = $ How will $f$ expand or contract space around the point $(\pi, \pi)$ ? Choose 1 answer: Choose 1 answer: (Choice A) A Leave it the same (Choice B) B Expand it finitely (Choice C) C Contract it finitely (Choice D) D Contract it infinitely
Explanation: The Jacobian determinant is the determinant of the Jacobian matrix. It represents the factor by which the transformation $f$ expands or contracts volume around a certain input. $\begin{aligned} |J(f)| &= \det \left( \begin{bmatrix} -\sin(x) & -\cos(y) \\ \\ \cos(x) & \sin(y) \end{bmatrix} \right) \\ \\ &= -\sin(x)\sin(y) + \cos(x)\cos(y) \end{aligned}$ If we evaluate $|J(f)|$ at $(\pi, \pi)$, we get $1$. Because the Jacobian determinant here has an absolute value equal to $1$, we can conclude that $f$ will leave the space around $(\pi, \pi)$ the same, not expanding nor contracting it. To recap, the Jacobian determinant of $f$ is $-\sin(x)\sin(y) + \cos(x)\cos(y)$, and $f$ will leave the space around the point $(\pi, \pi)$ the same, not expanding nor contracting it.